All fractions which can be written simultaneously in the forms $\frac{7k-5}{5k-3}$ and $\frac{6l-1}{4l-3}$

Question

Find all fractions which can be written simultaneously in the forms $\frac{7k-5}{5k-3}$ and $\frac{6l-1}{4l-3}$

for some integers $k,l$.

Please check my answer and tell me is correct or not....

$$\frac{43}{31},\frac{31}{27},1,\frac{55}{39},\frac{5}{3},\frac{61}{43},\frac{19}{13},\frac{13}{9}$$

Answer 1

Suppose there is integer $p$ which can be written as $\frac{6l-1}{4l-3}$ and $\frac{7k-5}{5k-3}$.

$$p= \frac{6l-1}{4l-3} =\frac{7k-5}{5k-3}$$

$$\implies kl+8k+l=6$$

$$\implies(k+1)l=(6-8k)\implies l=\frac{-2(4k-3)}{(k+1)}$$.

Which gives following integer solutions:

$(k,l)=(-15,-9),(-8,-10),(-3,-15),(-2,-22),(0,6),(1,-1),(6,-6),(13,7)$. These all sets of values will give you a new such number. I shall let you conclude now.

Answer 2

We have that $$\frac{7k-5}{5k-3}=\frac{6l-1}{4l-3}\iff kl+8k+l=6.$$ That is, if $k\ne -1,$

$$l=2\frac{3-4k}{k+1}=-2\left(4-\frac{7}{k+1}\right)=-8+\frac{14}{k+1}.$$ Since $l$ has to be an integer $k+1$ must divide $14.$ So, we have that $k\in\{-15,-8,-3,-2,0,1,6,13\}.$

Note that $k\ne -1$ since if $k=-1$ the equation $kl+8k+l=6$ gives $-8=6$ which doesn't hold.

Answer 3

$$\frac{7k-5}{5k-3}=\frac{6l-1}{4l-3}$$

$$28kl-20l-21k+15=30kl-18l-5k+3$$

$$2kl+2l+16k-12=0$$

$$kl+l+8k-6=0$$

either:

$$l(1+k)=2(3-4k)$$ so $$l=2\frac{3-4k}{1+k}$$

or:

$$k(l+8)=6-l$$

so

$$k=\frac{6-l}{l+8}$$

Let's go with this second one to complement the other answer.

Then $$k=-\frac{l-6}{l+8}=-\left(\frac{l+8-14}{l+8}\right)=-\left(1-\frac{14}{l+8}\right)$$

We want $l+8=\pm(1,2,7,14)$

So $l=-7,-9,-6,-10,-1,-15,6,-22$ or more nicely ordered $$l=-22,-15,-10,-9,-7,-6,-1,6$$

The pairs are then $(l,k)=$ $(-22,-2)$, $(-15,-3)$, $(-10,-8)$, $(-9,-15)$, $(-7,13)$, $(-6,6)$, $(-1,1)$, $(-22,-2)$, $(6,0)$, or that the fractions are

$$\frac{19}{13},\frac{13}{9},\frac{61}{43},\frac{55}{39},\frac{43}{31},\frac{37}{27},1,\frac{5}{3}.$$